Problem Solving

Rahul@gurome wrote:

gmatusa2010 wrote:The ratio of two positive numbers is 3 to 4. If k is added to each number the new ratio will be 4 to 5, and the sum of the numbers will be 117. What is the value of k?

1,
13,
14,
18,
21


I don't get this problem, isn't the 4 to 5 ratio useless information? Can't you solve this fine without that info. Can someone check my work. Also ambiguous with the sum of the numbers will be 117. My first read, I thought it was Old number 1, Old number 2, and K. But its not, according to the solution. Its actually new number 1 + new number 2= 117.

X/Y=3/4, X+K+Y+K=117 , X+Y+2k=117

Since, X+Y are multiples of 7, I'm going to re-write the problem as 7Q+2K=117
7Q= (117-2K) has to be divisible by 7. Number testing, yields only 13.

Everything is fine with your approach except a single flaw in your logic!
As you have found out three equations in three unknowns, which are

• (1) 4x = 3y
  (2) 5*(x + k) = 4*(y + k)
  (3) (x + y + 2k) = 117

Now the values of x or y or k can only be uniquely obtained if and only if we consider all the three equations together. Otherwise if we take any two of them, we cannot solve them for sure.

For example, in your case you have considered 1st and 3rd equation only.
Let's see what we can achieve with these two equations. From the first equation we have y = (4x/3). Replace y with this value in the third equation. Then we have, (x + 4x/3 + 2k) = 117. Thus,

• k = 117 - (7x/3)

As we have expected we cannot uniquely determine the value of k as it depends upon x.

Where have gone wrong then?
You've done a very common assumption which lead you in the wrong way! What is that? You assumed x and y to be integer and thus said that (x + y) is divisible by 7! In the question they were declared as numbers only! So basically you are assuming a new information and discarding a given information!

Conclusion: "The ratio of two positive numbers is 3 to 4 doesn't mean sum of the numbers is divisible by 7! For example 3/5 and 4/5 has a ratio 3 to 4, but their sum 7/5 is not divisible by 7!